Chapter5-Aircraft Engine Inlets and Nozzles

Ex1-pg251

In [10]:
#calcualte overspeed mach no
print("Example 5.1")
Md=1.5
##From isentropic table,
gm=1.4 ##gamma
A=1.176 ##A=A1/Ath=A1/Acr
##for same A, from isentropic table for M<1
My=0.61
##for My=0.61, from normal shock table
Mx=1.8
Mos=Mx
print'%s %.1f %s'%("Overspeed Mach no.",Mos,"")

Example 5.1
Overspeed Mach no. 1.8

Ex2-pg252

In [11]:
import math
#calculate contractio ratio and the maximum pressure recovery
print("Example 5.2")
Md=2.65
Mx=Md
##for Mx=2.65, from normal shock table 
My=0.4996
M1=My
##from isentropic table for M1=0.5, 
A=1.34
##for Md=2.65, from isentropic table (A=A1/Acr)
A1=3.036
Af=A1/A
##from isentropic table Af, 
Mth=2.35
##for Mth=2.35, from normal shock table
p=0.5615 ##p=pty/ptx
print'%s %.2f %s'%("Maximum total pressure recovery:",p,"")

Example 5.2
Maximum total pressure recovery: 0.56

Ex3-pg253

In [13]:
#calculate inlet design contraction ratio and throat mach no
print("Example 5.3")
Md=3.3 ##from isentropioc  table 
A=5.629 ## A=A1/Acr=A1/Ath
Mx=Md ##from normal shock table 
My=0.4596
M1=My
##from isentropic table 
A11=1.425
pt=((1./A11-1./A)/(1./A))*100.
Af=A/A11
##for Af=3.95, from isentropic table for M>1
M1th=2.95
print'%s %.2f %s'%("Inlet design contraction ratio A1/Ath:",A," ")
print'%s %.2f %s'%("The % opening of the throat:",pt," ")
print'%s %.3f %s'%("Throat Mach no.:",M1th,"")

Example 5.3
Inlet design contraction ratio A1/Ath: 5.63  
The % opening of the throat: 295.02  
Throat Mach no.: 2.950

Ex4-pg256

In [14]:
import math
#calculate inlet pressure recovery with the shock at the lip
print("Example 5.4")
M0=1.4
##from normal shock table 
p=0.9582 ##p=pt2/pt0
M1=M0
##from isentropic table:
A=1.115 ##A=A1/Acr
A11=1.1 ##A11=Ax/A1
Af=A11*A
##from normal shock table for M>1
Mx=1.56
##from normal table
p1=0.91 ##p=pt2/pt0
p2=p
print'%s %.2f %s'%("(a)The best backpressure :",p,"")
print'%s %.3f %s'%("(b)The supercritical mode inlet total pressure recovery:",p1,"")
print'%s %.2f %s'%("(c)Inlet pressure recovery in subcritical mode with 10% spillage:",p2,"")

Example 5.4
(a)The best backpressure : 0.96 
(b)The supercritical mode inlet total pressure recovery: 0.910 
(c)Inlet pressure recovery in subcritical mode with 10% spillage: 0.96

Ex5-pg257

In [15]:
#calculate total pressure recovery of this inlet 
import math
print("Example 5.5")
##th=theta and b=beta.
gm=1.4 ##gamma
##OBLIQUE SHOCK 1
M0=2.
th=8. ##degree
##from theta-beta-M chart,
b1=37. ##degree
Mn1=M0*math.sin(b1/57.3)
p1=0.993 ##p=pt2/pt1
Mn2=((2.+(gm-1.)*Mn1**2.)/(2.*gm*Mn1**2.-(gm-1.)))**(1/2.)
M2=Mn2/math.sin(b1-th/57.3)
##OBLIQUE SHOCK 2
M0=M2
th=12.
##from oblique shock chart,
b2=48.7
Mn1=M0*math.sin(b2/57.3)
p2=0.978
Mn2=((2.+(gm-1.)*Mn1**2)/(2.*gm*Mn1**2.-(gm-1.)))**(1/2.)
M3=Mn2/math.sin(b1-th/57.3)
##NORMAL SHOCK
M0=M3
b3=90.
pNS=0.977

Po=p1*p2*pNS
print'%s %.3f %s'%("Total pressure recovery:",Po,"")

Example 5.5
Total pressure recovery: 0.949

Ex6-pg271

In [16]:
import math
#calculate percent increase in gross thurst 
print("Example 5.6")
M9=1. ## Mach no.
p=1/8. ##p=p0/pt7
gm=1.3 ##gamma
V9cd=(2.*(1.-p**((gm-1.)/gm)))**(1/2.)
px=p*((gm+1.)/2.)**(gm/(gm-1.))
V9c=(2.*(gm-1.)/(gm+1.))**(1/2.)
FR=(V9cd/V9c)/(1.+(1.-px)/gm)
pr=(FR-1.)*100./1.
print'%s %.3f %s'%("% increase in gross thrust:",pr," ")

Example 5.6
% increase in gross thrust: 7.304

Ex7-pg273

In [18]:
#calculate velocites at various point and coefficent
import math
print("Example 5.7")
p98=0.95 ##p98=pt9/pt8
p87=0.98 ##p98=pt8/pt7
p70=8. ##p70=pt7/pt0
p97=8. ##p97=pt9/pt7
Cp=1243.7 ##specific heat in J/kg.K
gm=1.3 ##gamma
Tt9=900. ##Total temp. of the gas entering a C-D nozzle
Tt7=Tt9
p90=1. ##p90=p9/p0
p99=p98*p87*p70*p90 ##p99=pt9/p9
M9=(2./(gm-1.)*(p99**((gm-1.)/gm)-1.))**(1/2.) ##exit mach no.
T9=Tt9/(1.+(gm-1.)*M9**2/2.) ##The nozzle exit static temp.
a9=((gm-1.)*Cp*T9)**(1/2.) ##speed of sound in exit plane
V9=a9*M9 ##exit velocity
V9s=(2.*Cp*Tt7*(1.-p97**-((gm-1.)/gm)))**(1/2.)
p89=p87*p70*p90 ##p89=pt8/p9
V9i=(2.*Cp*Tt7*(1.-p89**-((gm-1.)/gm)))**(1/2.)
Cv=V9/V9i
print'%s %.1f %s'%("(a)V9 in",V9," m/s:")
print'%s %.1f %s'%("(b)V9s in",V9s," m/s:")
print'%s %.1f %s'%("(c)V9i in ",V9i,"m/s:")
print'%s %.3f %s'%("(d)The velocity coefficient Cv:",Cv,"")

Example 5.7
(a)V9 in 911.1  m/s:
(b)V9s in 923.7  m/s:
(c)V9i in  920.2 m/s:
(d)The velocity coefficient Cv: 0.990

Ex8-pg275

In [1]:
%matplotlib inline
#plot the graphs
print "Example 5.8"
#calculate and graph the divergnece correction factor 
import math
import numpy
import matplotlib
from matplotlib import pyplot
alfa=0 #alfa=cone half angle
dx=numpy.linspace(0,44,146)
x=numpy.zeros(146)
count=0;
for alfa in dx:
	Ca=(1+math.cos(alfa*math.pi/180.))/2.; #Flow angularity loss coefficient
	x[count]=Ca;
	count=count+1;
#disp(Ca,"Divergence correction factor Ca:")

pyplot.plot(dx,x)

pyplot.title("Flow convergence loss in a conical nozzle")
pyplot.xlabel("Cone half-angle")
pyplot.ylabel("Flow angularity loss coefficient")

Example 5.8
Out[1]:
<matplotlib.text.Text at 0x5989a50>

Ex9-pg276

In [2]:
print "Example 5.9"
#plot the graphs
import math
import numpy
import matplotlib
from matplotlib import pyplot
%matplotlib inline
alfa=0.1
dx=numpy.linspace(0.1,44,88)
x=numpy.zeros(88)
g1=numpy.zeros(88)
count=0
g2=numpy.zeros(88)
gc1=0;
for alfa in dx:
	Ca=(math.sin(alfa*math.pi/180.))/(alfa*math.pi/180.)
	Cac=(1+math.cos(alfa*math.pi/180.))/2.
	x[count]=Ca
	count=count+1;
	g1[gc1]=Cac;
	gc1=gc1+1;


pyplot.plot(dx,g1)
pyplot.plot(dx,x)
pyplot.legend(["Conical","2D-CD"])
pyplot.xlabel("Divegent flap angle or Cone half-angle(degree)")
pyplot.ylabel("Flow angularity loss coefficient")
pyplot.title("Divergent loss of a conical nozzle and a 2D-CD nozzle")

Example 5.9
Out[2]:
<matplotlib.text.Text at 0x5b714d0>

Ex10-pg282

In [3]:
print ("Example 5.10")
%matplotlib inline
#plot the graphs
import numpy
import matplotlib
from matplotlib import pyplot
p=0.96 #p=p't8/pt8
f=0.02
fAB=0.04

z0=numpy.linspace(0.45,0.63,7)
gmr=1.3/1.33 #gm=gm/gm' gm=gamma
gm=1.33
gm1=1.3
tlAB=7.
tl=6.
i=0;
z1=numpy.linspace(7,9,3)
for tlAB in z1:
    tt=6.5
    g1=numpy.zeros(7)
    gc1=0;
    for tt in z0:
        A=(1+f+fAB)/(1+f)*((gmr)**(1./2))*1/p*((tlAB/(tl*tt))**(1./2))*((((gm1+1)/2.)**((gm1+1)/(2*(gm1-1))))/(((gm+1)/2.)**((gm+1)/(2.*(gm-1)))))
        g1[gc1]=A
        gc1=gc1+1;
    number=0;
    pyplot.plot(z0,g1)
    i=i+1;
    pyplot.xlabel("Turbine expansion parameter")
    pyplot.ylabel("A8-AB-ON/A8-AB-OFF")
    pyplot.title("Nozzle throat area variation with ")
    pyplot.legend(["tau(AB)=7","tau(AB)=8","tau(AB)=9"])

Example 5.10

Ex12-pg293

In [4]:
print ("Example 5.12")
#plot the graphs
%matplotlib inline
import numpy
import matplotlib
from matplotlib import pyplot
import warnings
warnings.filterwarnings('ignore')
gm=1.1
M0=2.5
g1=numpy.zeros(40)

z0=numpy.linspace(0,4,40)
i=1;
z1=numpy.linspace(1.1,1.4,4)
for gm in z1:
    gc1=0;
    for M in z0:
		p0=(1+(gm-1)/2*(M**2))**(gm/(gm-1))
		p20=.4*p0-.5*p0
		M=3
		p42=0.37
		NPR=p20*p42
		g1[gc1]=p0
		gc1=gc1+1;
		pyplot.plot(z0,g1)
		pyplot.title("Total-to-static pressure ratio")
		pyplot.xlabel("Flight Mach no. (M0)")
		pyplot.ylabel("pt0/p")
		pyplot.legend(["gamma=1.1","gamma=1.2","gamma=1.3","gamma=1.4"])
		i=i+1;

Example 5.12

Ex13-pg296

In [5]:
print "Example 5.13"
#plot the graphs
%matplotlib inline
import numpy
from numpy import linspace
import matplotlib
from matplotlib import pyplot
#T=Th/Tc
z0=numpy.linspace(0,8,160)
i=0
z1=numpy.linspace(1,4.5,7)
for T in z1:
	g1=numpy.zeros(160);
	gc1=0;
	for alfa in z0:
		FR=((1+alfa)**(1./2)*(T+alfa)**(1./2))/(T**(1./2)+alfa)
		g1[gc1]=FR
		gc1=gc1+1;
	number=0;
	pyplot.plot(z0,g1)
	i=i+1;
	pyplot.xlabel("Bypass ratio(alfa)")
	pyplot.ylabel("Ratio of mixed to seperate-flow turbofan engines gross thrust")
	pyplot.legend("T(hot)/T(cold)=1.5","T(hot)/T(cold)=2","T(hot)/T(cold)=2.5 so on")
	pyplot.title("Ideal gross thrust gain with a perfect mixer")

Example 5.13